1. Introduction
Let H be a Hilbert space of analytic functions on the unit disk D. For an analytic function ψ on D, we can define the multiplication operator Mψ:f→ψf, f∈H. For an analytic self-mapping φ of D, the composition operator Cφ defined on H as Cφf=f∘φ, f∈H. These operators are two classes of important operators in the study of operator theory in function spaces [1–3]. Furthermore, for ψ and φ, we define the weighted composition operator Cψ,φ on H as
(1.1)Cψ,φ:f⟶ψ(f∘φ), f∈H.

Recently, the boundedness, compactness, norm, and essential norm of weighted composition operators on various spaces of analytic functions have been studied intensively, see [4–9] and so on. In this paper, we characterize bounded Fredholm weighted composition operators on Dirichlet space of the unit disk.

Recall the Dirichlet space 𝒟 that consists of analytic function f on D with finite Dirichlet integral:
(1.2)D(f)=∫D|f′|2dA<∞,
where dA is the normalized Lebesgue area measure on D. It is well known that 𝒟 is the only möbius invariant Hilbert space up to an isomorphism [10]. Endow 𝒟 with norm
(1.3)‖f‖=(|f(0)|2+D(f))1/2, f∈D.𝒟 is a Hilbert space with inner product
(1.4)〈f,g〉=f(0)g(0)¯+∫Df′(z)g′(z)¯dA(z), f,g∈D.
Furthermore 𝒟 is a reproducing function space with reproducing kernel
(1.5)Kλ(z)=1+log11-λ-z, λ,z∈D.

Denote ℳ={ψ:ψ is analytic on D, ψf∈𝒟 for f∈𝒟}. ℳ is called the multiplier space of 𝒟. By the closed graph theorem, the multiplication operator Mψ defined by ψ∈ℳ is bounded on 𝒟. For the characterization of the element in ℳ, see [11].

For analytic function ψ on D and analytic self-mapping φ of D, the weighted composition operator Cψ,φ on 𝒟 is not necessarily bounded. Even the composition operator Cφ is not necessarily bounded on 𝒟, which is different from the cases in Hardy space and Bergman space. See [12] for more information about the properties of composition operators acting on the Dirichlet space.

The main result of the paper reads as the following.

Theorem 1.1.
Let ψ and φ be analytic functions on D with φ(D)⊂D. Then Cψ,φ is a bounded Fredholm operator on 𝒟 if and only if ψ∈ℳ, bounded away from zero near the unit circle, and φ is an automorphism of D.

If ψ(z)=1, then the result above gives the characterization of bounded Fredholm composition operator Cφ on 𝒟, which was obtained in [12].

As corollaries, in the end of this paper one gives the characterization of bounded invertible and unitary weighted composition operator on 𝒟, respectively. Some idea of this paper is derived from [4, 13], which characterize normal and bounded invertible weighted composition operator on the Hardy space, respectively.

2. Proof of the Main Result
In the following, ψ and φ denote analytic functions on D with φ(D)⊂D. It is easy to verify that ψ∈𝒟 if Cψ,φ is defined on 𝒟.

Proposition 2.1.
Let Cψ,φ be a bounded Fredholm operator on 𝒟. Then ψ has at most finite zeroes in D and φ is an inner function.

Proof.
If Cψ,φ is a bounded Fredholm operator, then there exist a bounded operator T and a compact operator S on 𝒟 such that
(2.1)T(Cψ,φ)*=I+S,
where I is the identity operator.

Since
(2.2)(Cψ,φ)*Kw(z)=〈Cψ,φ*Kw,Kz〉=〈Kw,Cψ,φKz〉=〈Kw,ψKz∘φ〉=ψ(w)Kz(φ(w))¯=ψ(w)¯Kφ(w)(z),
we have
(2.3)‖T‖|ψ(w)|‖Kφ(w)‖‖Kw‖≥‖T(Cψ,φ)*kw‖≥‖kw‖-‖Skw‖=1-‖Skw‖,
where kw=Kw/∥Kw∥ is the normalization of reproducing kernel function Kw.

Since S is compact and kw weakly converges to 0 as |w|→1, ∥Skw∥→0 as |w|→1. It follows that there exists constant r, 0<r<1, such that ∥Skw∥<1/2 for all w with r<|w|<1. Inequality (2.3) shows that
(2.4)|ψ(w)|‖Kw‖≥12‖T‖‖Kφ(w)‖, r<|w|<1,
which implies that ψ has no zeroes in {w∈D, r<|w|<1}, and, hence, ψ has at most finite zeroes in {w∈D, |w|≤r}.

Since kw weakly converges to 0 as |w|→1, 〈ψ,kw〉→0 as |w|→1, that is,
(2.5)ψ(w)‖Kw‖⟶0, |w|⟶1.
It follows from (2.4) that ∥Kφ(w)∥=(1+log (1/(1-|φ(w)|2)))1/2→∞ and hence |φ(w)|→1 as |w|→1, that is, φ is an inner function.

For the proof of the following lemma, we cite Carleson's formula for the Dirichlet integral [14].

Let f∈𝒟, f=BSF be the canonical factorization of f as a function in the Hardy space, where B=∏j=1∞(a-j/|aj|)((aj-z)/(1-a-jz)), is a Blaschke product, S is the singular part of f and F is the outer part of f. Then
(2.6)D(f)=∫T∑n=1∞Pαn(ξ)|f(ξ)|2|dξ|2π+∬T2|ζ-ξ|2|f(ξ)|2dμ(ζ)|dξ|2π+∬T(e2u(ζ)-e2u(ξ))(u(ζ)-u(ξ))|ζ-ξ|2|dζ|2π|dξ|2π,
where T is the unit circle, u(ξ)=log |f(ξ)|, Pα(ξ) is the Poisson kernel, and μ is the singular measure corresponding to S.

Lemma 2.2.
Let Cψ,φ be a bounded operator on 𝒟, ψ=BF with B a finite Blaschke product. Then CF,φ is bounded.

Proof.
Let MB be the multiplication operator on 𝒟. Then Cψ,φ=MBCF,φ. Since B is a finite Blaschke product, by the Carleson's formula, we have
(2.7)D(ψ(f∘φ))=D(BF(f∘φ))≥D(F(f∘φ)), f∈D.

Since ∥f∥2=|f(0)|2+D(f), f∈𝒟, by the inequality above it is easy to verify that CF,φ is bounded on 𝒟 if Cψ,φ is bounded.

Lemma 2.3.
Let F be an analytic function on D with zero-free. If CF,φ is a bounded Fredholm operator on 𝒟, then φ is univalent.

Proof.
If φ(a)=φ(b) for a,b∈D with a≠b, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {an} and {bn} in D which is disjoint such that φ(an)=φ(bn). Hence,
(2.8)(CF,φ)*(KanF(an)¯-KbnF(bn)¯)=0,
which contradicts to that kernel of (CF,φ)* is finite dimensional.

Corollary 2.4.
If Cψ,φ is a bounded Fredholm operator on 𝒟, then φ is an automorphism of D and ψ∈ℳ.

Proof.
By Proposition 2.1, ψ has the factorization of BF with B a finite Blaschke product and F zero free in D. By Lemma 2.2, CF,φ is a bounded operator on 𝒟. Since Cψ,φ=MBCF,φ and MB is a Fredholm operator, CF,φ is a Fredholm operator also. By Proposition 2.1 and Lemma 2.3, φ is an univalent inner function, it follows from [1, Corollary 3.28] that φ is an automorphism of D.

Since Cψ,φCφ-1=Mψ, Mψ is a bounded multiplication operator on 𝒟, which implies that ψ∈ℳ.

The following lemmas is well-known. It is easy to verify by the fact Mψ*Kw=ψ(w)¯Kw also.

Lemma 2.5.
Let ψ∈ℳ. Then Mψ is an invertible operator on 𝒟 if and only if ψ is invertible in ℳ.

Lemma 2.6.
Let ψ∈ℳ. Then Mψ is a Fredholm operator on 𝒟 if and only if ψ is bounded away from the unit circle.

Now we give the proof of Theorem 1.1.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>.
If Cψ,φ is a bounded Fredholm operator on 𝒟, by Corollary 2.4, ψ∈ℳ and φ is an automorphism of D. Since Cφ is invertible, Mψ is a Fredholm operator. So ψ is bounded away form the unit circle follows from Lemma 2.6.

On the other hand, if ψ∈ℳ and bounded away from the unit circle, then Mψ is a bounded Fredholm operator on 𝒟. If φ is an automorphism of D, then Cφ is invertible. Hence Cψ,φ=MψCφ is a bounded Fredholm operator on 𝒟.

As corollaries, in the following, we characterize bounded invertible and unitary weighted composition operators on 𝒟.

Corollary 2.7.
Let ψ and φ be analytic functions on D with φ(D)⊂D. Then Cψ,φ is a bounded invertible operator on 𝒟 if and only if ψ∈ℳ, invertible in ℳ, and φ is an automorphism of D.

Proof.
Since a bounded invertible operator is a bounded Fredholm operator, the proof is similar to the proof of Theorem 1.1.

Corollary 2.8.
Let ψ and φ be analytic functions on D with φ(D)⊂D. Cψ,φ is a bounded operator on 𝒟. Then Cψ,φ is a unitary operator if and only if ψ is a constant with |ψ|=1 and φ is a rotation of D.

Proof.
If Cψ,φ is a unitary operator, then it must be an invertible operator. By Corollary 2.7, φ is an automorphism of D and ψ is invertible in ℳ.

Let n be nonnegative integer, en(z)=zn, z∈D. A unitary is also an isometry, so we have
(2.9)‖ψ‖=‖Cψ,φe0‖=‖e0‖=1,(2.10)‖ψφn‖=‖Cψ,φen‖=‖en‖=n, n≥1.

Let α∈D such that φ(α)=0. Since φ is an automorphism of D, φn is a finite Blaschke product with zero α of order n. By Carleson’s formula for Dirichlet integral, we have
(2.11)D(ψφn)=n∫TPα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ).
Hence,
(2.12)n=‖ψφn‖2=|ψ(0)φ(0)n|2+D(ψφn)=|ψ(0)φ(0)n|2+n∫TPα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ), n≥1.
That is,
(2.13)1=|ψ(0)φ(0)n|2n+∫TPα(ξ)|ψ(ξ)|2|dξ|2π+D(ψ)n, n≥1.
Let n→∞, then 1=∫TPα(ξ)|ψ(ξ)|2(|dξ|/2π).

By (2.12), we have D(ψ)=0 and |ψ(0)φ(0)|=0. By (2.9), we obtain ψ is a constant with |ψ|=1, which implies that φ(0)=0, that is, φ is a rotation of D.

The sufficiency is easy to verify.

Remark 2.9.
The key step in the proof of the main result is to analyze zeros of the symbol ψ and univalency of φ. The following result pointed out by the referee gives a simple characterization of the symbols ψ and φ for the bounded Fredholm operator Cψ,φ on 𝒟.

Proposition 2.10.
Let ψ and φ be analytic functions on D with φ(D)⊂D. Cψ,φ is a bounded Fredholm operator on 𝒟. Then ψ has only finitely many zeros in D and φ is univalent.

Proof.
If ψ(a)=0 for a∈D, then Cψ,φ*Ka=ψ(a)¯Kφ(a)=0, which implies that Ka is in the kernel of Cψ,φ*. Thus if ψ had infinitely many zeros, the kernel of Cψ,φ* would be infinite dimensional and hence this operator would not be Fredholm.

If φ(a)=φ(b) for a,b∈D with a≠b, by a similar reasoning as [1, Lemma 3.26], there exist infinite sets {an} and {bn} in D which is disjoint such that φ(an)=φ(bn). Since ψ has only finitely many zeros in D, we can choose infinitely many an and bn such that ψ(an)≠0, ψ(bn)≠0. Hence,
(2.14)(Cψ,φ)*(Kanψ(an)¯-Kbnψ(bn)¯)=0.
Since Cψ,φ is a Fredholm operator, φ must be univalent.